Question

lim h--> 0

(sin (pi/4+h) - sin pi/4)/ h

Answer 1

do you know how to take the derivative of sin(x) ?
or do you have to evaluate this limit without using the derivative?

Answer 2

you could use the trig identity
sin(A+B)= sin(A) cos(B) + cos(A) sin(B)

you also need to know lim h->0 cos(h)= 1
and lim h->0 sin(h) = h

Answer 3

because your limit is the definition of the derivative of sin(x) evaluated at x=pi/4 you could just do
d/dx sin(x) = cos(x)
and then evaluate cos(pi/4)

Answer 4

Use sin(A+B)=sinAcosB+cosAsinB
(sin(pi/4+h) - sin(pi/4))/h
=(sin(pi/4)cos(h)+cos(pi/4)sin(h)-sin(pi/4))/h
=(sqrt2/2)cos(h) + (sqrt(2)/2)*sin(h)-(sqrt2/2)/h
this is where im stuck

Answer 5

the answer should be 1/sqrt2

Answer 6

you also need to know lim h->0 cos(h)= 1
and lim h->0 sin(h) = h

Answer 7

but idk what to do next after

(sqrt2/2)cos(h) + (sqrt(2)/2)*sin(h)-(sqrt2/2)/h

Answer 8

cos(h) -->1 and sin(h)-->h
so you have
\[ \frac{(\sqrt{2}/2)1 + (\sqrt{2}/2)h-(\sqrt{2}/2)}{h} \]

Answer 9

yes

Answer 10

do you see sqr(2)/2 - sqr(2)/2 =0
leaving
\[ \frac{(\sqrt{2}/2)h}{ h} \]

Answer 11

yes i understand that
then you cross/divide off the H which will give 1

Answer 12

leaving
\[ \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \]

Answer 13

omg thank you som much =)

Answer 14

yw